2. Algebraic Categories  Definition 2.1. An algebraic category is a concrete category satisfied the following conditions:  (a) It has free objects.  (b) Any bijective morphism is an isomorphism.  (c) Any pair of parallel morphisms of objects has a surjective coequalizer.  Combining (1.9) and (1.10) we have the following variants of (2.1):  Proposition 2.2. A concrete category is an algebraic category iff the following conditions are satisfied:  (a) It has free objects.  (b) Any surjective morphism is a coequalizer.  (c) Any pair of parallel morphisms has a surjective coequalizer.  Proposition 2.3. A concrete category is an algebraic category iff the following conditions are satisfied:  (a) It has free objects.  (b) A function f: Y --> X of objects is a morphism if there is a surjective morphism g: Z --> Y such that fg is a morphism.  (c} Any surjective function from an object to a set has a surjective generic extension.  Proposition 2.4. (a) Any algebraic category A has a terminate object.  (b) An object in A is a terminate object iff its carrier is a singleton.  Proof. (a) First we note that A has an object whose carrier is a singleton. Let Z[Q] be the free object on a singleton Q. Let Z[Q] --> Q be the surjective function. By (1.9) Z[Q] --> Q has a surjective generic extension Q --> P. Then P is an object whose carrier is a singleton.  Next we show that for any object X there is a unique morphism to P. Consider the canonical morphism Z[X] --> X, whose composition with the surjective function X --> P is a morphism. By (1.9.b) this implies that X --> P is a morphism, which is obviouse unique as P is a singuleton.  (b) One direction has been noticed in (a). Suppose P' is any object whose carrier is a singuleton. Then the morphism from P' to any terminate object is bijective, therefore an isomorphism.  Suppose A is an algebraic category.  Proposition 2.5. Any morphism f: Y --> X of objects factors uniquely as a surjective morphism followed by an injective morphism.  Proof. Let f(Y) be the image of the set Y under the function f. Let q: Y --> f(Y) be the surjective function and e: f(Y) --> X the inclusion. Let t: f(Y) --> B be the surjective generic extension of q (see (2.3.c)). As f = eq is a morphism, by the definition of a generic extension there is a unique morphism h: B --> X such that e = ht. This implies that the function t is injective, therefore it is bijective. Since B is only defined up to isomorphism, and f(Y) carries a A -structure such that t is an isomorphism of objects by (1.2), we may assume B = f(Y) . This shows that f factors as a surjective morphism q: Y --> f(Y) followed by an injective morphism f(Y) --> X of objects.  Definition 2.6. (a) By a subobject of X we mean an object (V, TV) such that V is a subset of X, and the inclusion V --> X is a morphism of objects, called the inclusion of a subobject; TV is a subobjectic structure on V.  (b) A subset of an object X is called closed if it is the carrier of a subobject.  Proposition 2.7. A subset of X is closed iff it is the image of a morphism to X .  Proof. This follows from (2.5).  Proposition 2.8. (a) Suppose (U, TU) and (V, TV) are two subobjects of an object X and V U. Then the inclusion V --> U is a morphism.  (b) Suppose (V, TV) and (V, SV) are two subobjects of an object X with the same carrier. Then TV = SV .  (c} If f: Y --> X is a morphism of objects and U is a subobject of X containing the image f(Y) of Y. Then the morphisms f factors through the inclusion U --> X of the subobject U.  Proof. (a) Consider the morphism d: Z[V] --> X, s: Z[V] --> (V, TV) and t: Z[V] --> (U, TU) determined by the inclusions V --> X, the identity V --> V, and the inclusion V --> U of sets respectively. Then d = ee's = e't where e: U --> X and e': V --> U are the inclusions of sets. Since e is injective, we have e's = t. Since s is surjective and t is a morphism, by (2.3.b) the inclusion e' is a morphism.  (b) follows from (a) and (1.2).  (c) follows from (a) and (2.1).  Remark 2.9. It follows from (2.8.b) that a subobject is uniquely determined by its carrier. This justifies the common practice in algebra that the carrier of a subobject is simply called a subobject (i.e. subalgebra).  If U is a subset of an object X, let g: Z[U] --> X be the morphism determined by the inclusion U --> X. We shall write G(U) for the image g(Z[U]), which is a subset of X.  Proposition 2.10. (a) A subset U of an object X is closed iff U = G(U).  (b) G(G(U)) = G(U) for any subset U of X.  (c) U G(U) for any subset U of X.  (d) If U V are two subsets of X then G(U) G(V).  (e) G(U) is the smallest closed subset of X containing U.  (f) Any intersection of closed subset is closed.  Proof. (a) If U = G(U) then U is the image of the morphism g: Z[U] --> X, so it is closed by (2.7). Conversely, assume U = f(Y) where f: Y --> X is a morphism of objects. Let d: Z[f(Y)] --> X be the morphism determined by the inclusion d': f(Y) --> X. Let s: f(Y) --> Y be a section of f': Y --> f(Y). Let s': Z[f(Y)] --> Y be the morophism determined by s. Then d = fs'. Thus d(Z[f(Y)]) = f(s'(Z[f(Y)])) f(Y). Thus G(f(Y)) = f(Y).  (b) follows from (a) and (c) is obvious.  (d) Suppose e: U --> V and e': V --> X are the inclusions. Let d: Z[U] --> Z[V] be the morphism determined by the inclusion, and let d': Z[V] --> X be the morphisms determined by e'. Then G(U) = d'd(Z[U]) d'(Z[V]) = G(V).  (e) If W is an algebraic subset containing U then by (c) we have G(W) G(U). This shows that G(U) is the smallest closed subset containing U.  (f) Suppose {Ui} is a set of subalgebras. Let U = i Ui. Then U Ui implies that G(U) G(Ui) = Ui as each Ui is closed. Thus U G(U)  i Ui = U, i.e. U = G(U). Thus U is closed.  Definition 2.11. (a) G(U) is called the closed subset generated by U.  (b) If X = G(U) then we say that U is a set of generatorsof X.  Example 2.11.1. If Z[S] is a free object on a set S, then S is a set of generators of Z[S].  Proposition 2.12. (a) Suppose f: Y --> X and g: Y --> Z are two surjective morphisms. If there is a bijection h: X --> Z such that g = hf, then h is an isomorphism.  (b} If t: Y --> T is a surjective function, there is at most one structure on T such that t is a morphism.  Proof (a) Note that g and G-1 are morphisms by (2.3.b).  (b) follows from (a).  Example 2.11.1. We verify that the concrete category Set of sets is a free algebraic category.  (a) Clearly any set is free on itself.  (b) Any bijective function is an isomorphism.  (c) Given two functions f, g: Y --> X in Set, let ~ be the smallest equivalence relation on X such that f(y) ~ g(y) for all y in Y. Then the natural map q: X --> X/~, which assigns to each x in X the equivalence relation class to which x belongs, is a surjective coequalizer of f and g.        [Next Section][References][Notations][Home] 